The distance traveled, in feet of a ball droped from a tall building is modeled by the equation d(t) = 16t^2 where d equals the distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 2 to t = 5 represent?

Respuesta :

That represents the avg acceleration of the ball from 2-5sec

[tex] \bf slope = m = \cfrac{rise}{run} \implies
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------\\\\
d(t)= 16t^2 \qquad
\begin{cases}
t_1=2\\
t_2=5
\end{cases}\implies \cfrac{d(5)-d(2)}{5-2}\implies \cfrac{400-64}{5-2}
\\\\\\
\cfrac{336}{3}\implies 112 [/tex]


what does 112 ?


well, it means that from the 2nd second to the 5th second, on average the ball was dropping at an average of 112 feet per second.

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